218-5*(x+8)=\(2^2\)
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a.
$5^{75}=(5^5)^{15}=3125^{15}$
$7^{60}=(7^4)^{15}=2401^{15}$
Mà $3125> 2401$ nên $5^{75}> 7^{60}$
b.
$3^{21}=3.3^{20}=3.9^{10}$
$2^{31}=2.2^{30}=2.8^{10}< 3. 9^{10}$
$\Rightarrow 3^{21}> 2^{31}$

\(a)24\times(x-16)=12^2\\\Rightarrow 24\times(x-16)=144\\\Rightarrow x-16=144:24\\\Rightarrow x-16=6\\\Rightarrow x=6+16\\\Rightarrow x=22\\---\\b)(x^2-10):5=5\\\Rightarrow x^2-10=5\times5\\\Rightarrow x^2-10=25\\\Rightarrow x^2=25+10\\\Rightarrow x^2=35\\\Rightarrow x=\pm\sqrt{35}\\---\)
\(c)(5x+335):2=400\\\Rightarrow 5x+335=400\times2\\\Rightarrow 5x+335=800\\\Rightarrow 5x=800-335\\\Rightarrow 5x=465\\\Rightarrow x=465:5\\\Rightarrow x=93\\---\\d)63:(5x+4)=2^3-1\\\Rightarrow 63:(5x+4)=8-1\\\Rightarrow 63:(5x+4)=7\\\Rightarrow 5x+4=63:7\\\Rightarrow 5x+4=9\\\Rightarrow 5x=9-4\\\Rightarrow 5x=5\\\Rightarrow x=5:5\)
\(\Rightarrow x=1\)
\(Toru\)




Lời giải:
$a-11b+3c\vdots 17$
$\Rightarrow 2(a-11b+3c)\vdots 17$
$\Rightarrow 2a-22b+6c\vdots 17$
$\Rightarrow 2a-5b+6c-17b\vdots 17$
$\Rightarrow 2a-5b+6c\vdots 17$ (đpcm)

`#3107.101107`
a)
\(49\cdot52+48\cdot132-48\cdot83\\ =49\cdot52+48\cdot\left(132-83\right)\\ =49\cdot52+48\cdot49\\ =49\cdot\left(52+48\right)\\ =49\cdot100\\ =4900\)
b)
\(42\cdot53+47\cdot186-114\cdot47\\ =42\cdot53+47\cdot\left(186-114\right)\\ =42\cdot53+47\cdot72\\ =2226+3384\\ =5610\)
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Ta có:
5⁶⁰ⁿ = (5³)²⁰ⁿ = 125²⁰ⁿ
2¹⁴⁰ⁿ = (2⁷)²⁰ⁿ = 128²⁰ⁿ
3¹⁰⁰ⁿ = (3⁵)²⁰ⁿ = 243²⁰ⁿ
Do 125 < 128 < 243
125²⁰ⁿ < 128²⁰ⁿ < 243²⁰ⁿ
Vậy 5⁶⁰ⁿ < 2¹⁴⁰ⁿ < 3¹⁰⁰ⁿ
\(218-5\cdot(x+8)=2^2\\\Rightarrow 218-5\cdot(x+8)=4\\\Rightarrow 5\cdot(x+8)=218-4\\\Rightarrow 5\cdot(x+8)=214\\\Rightarrow x+8=214:5\\\Rightarrow x+8=42,8\\\Rightarrow x=42,8-8\\\Rightarrow x=34,8\)
⇒218-5*(x+8)=4
⇒5*(x+8)=218-4
⇒5*(x+8)=214
⇒x+8=214:5
⇒x+8=42,8
⇒x=34,8
⇒x